entropy change.ppt

Entropy

• Introduce the thermodynamic property called Entropy (S)

• Entropy is defined using the Clausius inequality

• Introduce the Increase of Entropy Principle which states that

– the entropy for an isolated system (or a system plus its surroundings) is always increases or, at best, remains the same.

– Second Law in terms of Entropy

• Learn to use the Entropy balance equation: entropy change = entropy transfer + entropy change.

• Analyze entropy changes in thermodynamic process and learn how to use thermodynamic tables

• Examine entropy relationships (Tds relations), entropy relations for ideal gases.

• Property diagrams involving entropy (T-s and h-s diagrams)

In this Chapter, we will:

Entropy (cont’d)

Since entropy is a thermodynamic property, it has fixed values at a fixed thermodynamic states. Hence, the change, S, is determined by the initial and final state. BUT..

The change is = only for a Reversible Process

1

2

T

S

2 2 2

2 1

1 1 1

2 1

2 1

1 2

0 From entropy definition

Q Q

dS= , 0

Therefore,

rev

rev rev _rev rev

rev

Q Q Q

T T T

Q Q

dS

T T T T

Q Q

dS S S S

T T

Q

S S S

T                _{ } _{ }        _{ }   _   _                     _   _{    }             _{   }  



























, This is valid for all processes

Q Q

, since = ,

T _rev T _irrev

Q

dS dS dS

T       _{ } _{ }    



Consider a cycle, where

Process 2-1 is reversible and 1-2 may or may not be

Increase of Entropy Principle

Implications:

•Entropy, unlike energy, is non-conservative since it is always increasing.

•The entropy of the universe is continuously increasing, in other words, it is becoming disorganized and is approaching chaotic.

• The entropy generation is due to the presence of irreversibilities. Therefore, the higher the entropy generation the higher the irreversibilities and, accordingly, the lower the efficiency of a device since a reversible system is the most efficient system.

• The above is another statement of the second law

Entropy change

Entropy Transfer

(due to heat transfer)

Entropy Generation

The principle states that for an isolated Or a closed adiabatic Or System + Surroundings

A process can only take place such that S_gen 0 where S_ge = 0 for a reversible process only And S_ge can never be les than zero.

Second Law & Entropy Balance

• Increase of Entropy Principle is another way of stating the Second Law of Thermodynamics:

Second Law : Entropy can be created but NOT destroyed

(In contrast, the first law states: Energy is always conserved)

•Note that this does not mean that the entropy of a system cannot be reduced, it can. • However, total entropy of a system + surroundings cannot be reduced

• Entropy Balance is used to determine the Change in entropy of a system as follows: Entropy change = Entropy Transfer + Entropy Generation where,

Entropy change = S = S₂ - S₁

Entropy Transfer = Transfer due to Heat (Q/T) + Entropy flow due to mass flow (m_is_i – m_es_e) Entropy Generation = S_gen 0

For a Closed System: S₂ - S₁ = Q_k /T_k + S_gen In Rate Form: dS/dt = Q_k /T_k + S_gen For an Open System (Control Volume):

Similar to energy and mass conservation, the entropy balance equations can be simplified Under appropriate conditions, e.g. steady state, adiabatic….

CV gen e

e i

i k

k

cv _{m s m s S}

T Q dt

dS

,

 

 

 

 

 

Entropy Generation Example

Show that heat can not be transferred from the low-temperature sink

to the high-temperature source based on the increase of entropy

principle.

Source

800 K

Sink

500 K

Q=2000 kJ

S(source) = 2000/800 = 2.5 (kJ/K) S(sink) = -2000/500 = -4 (kJ/K)

S_gen= S(source)+ S(sink) = -1.5(kJ/K) < 0

It is impossible based on the entropy increase principle

S_gen0, therefore, the heat can not transfer from low-temp. to high-temp. without external work input

• If the process is reversed, 2000 kJ of heat is transferred from the source to the sink, S_gen=1.5 (kJ/K) > 0, and the process can occur according to the second law

• If the sink temperature is increased to 700 K, how about the entropy generation? S(source) = -2000/800 = -2.5(kJ/K)

S(sink) = 2000/700 = 2.86 (kJ/K)

S_gen= S(source)+ S(sink) = 0.36 (kJ/K) < 1.5 (kJ/K)

Entropy Change

• Property diagrams (T-s and h-s diagrams)

– From the definition of the entropy, it is known that Q=TdS during a reversible process.

– Hence the total heat transfer during such a process is given by Q_reversible =  TdS

• Therefore, it is useful to consider the T-S diagram for a reversible process involving heat transfer

On a T-S diagram, the area under the process curve represents the heat transfer for a reversible process.

T

Example

Show a Carnot cycle on a T-S diagram and identify the heat transfer at both the high and low temperatures, and the work output from the cycle.

S

T

T_H

T_L

S₁=S₄ S₂=S₃

1

₂

3

4

A B

• 1-2, reversible isothermal heat transfer Q_H = TdS = T_H(S₂-S₁) area 1-2-B-A

• 2-3, reversible, adiabatic expansion isentropic process, S=constant (S₂=S₃)

• 3-4, reversible isothermal heat transfer Q_L = TdS = T_L(S₄-S₃), area 3-4-A-B

• 4-1, reversible, adiabatic compression isentropic process, S₁=S₄

Net work W

= Q

- Q

, is the area enclosed by 1-2-3-4, the

shaded area

Q

Q

Mollier Diagram

• Enthalpy-entropy diagram,

h-s

diagram: it is valuable in

analyzing steady-flow devices such as turbines,

compressors, etc.

•



h: change of enthalpy from energy balance (from the

first law of thermodynamics)

•



s: change of entropy from the second law ( a measure of

the irreversibilities during an adiabatic process)



s



h

h

TdS (

Gibbs

) Equations

• Since, the area under the T-s line is equal to the heat transfer for a reversible process, it would be useful to have a relationship between Temperature and Entropy to obtain the heat transfer,T.

• Such a relationship exists for a closed system containing a pure,

compressible substance undergoing a reversible process

dU = Q_rev - W_rev = TdS - PdV

=> TdS = dU + PdV or Tds = du + pdv ( per unit mass)

• This is the well-known Gibbs equation

• Alternatively, eliminate du by using the definition of enthalpy, h = u + pv => dh = du + pdv + vdp, => du + pdv = dh – vdp

Tds = du + pdv

Comments Regarding theTdS Equations

Equations relate the entropy change of a system to other

properties, such as enthalpy (h), internal energy (u), pressure

and volume.

• Hence, even though T-ds relations were derived for a

reversible process,

they are valid for any process, reversible

or irreversible

.

• They are applicable for closed or open systems.

Example – Application of Tds relations

• Consider steam undergoing a phase transition from liquid to vapor at a constant temperature of 20°C. Determine the entropy change s_fg=s_g-s_f using the Gibbs equations and compare the value to that read directly from the thermodynamic table.

)

(

)

(

1

,

f g

f g

f g

fg

v

v

T

P

u

u

T

s

s

s

dv

T

P

T

du

ds

















change

from

liquid

to

vapor

From table A-4:

T=20°C, P = 2.338 kPa, v_f = 0.001002(m3_{/kg), v}

g=57.79(m3/kg),

u_f=83.9(kJ/kg), u_g=2402.9(kJ/kg) Substituting into the Tds relation:

s_fg= (1/293)(2402.9-83.9) + (2.338/293)(57.79-0.001002) = 8.375(kJ/kg K)

Compares favorably with the tabulated value s_fg=8.3715(kJ/kg K)

Entropy Change of an Incompressible Substance

(YAC:6-8)

• For most liquids and all solids, the density does not change appreciably as pressure changes, hence dv  0.

• Gibbs equation states that Tds = du+pdv

 Tds = du where du = CdT, for an incompressible substance and C_p=C_v=C is a function of temperature only.

• Therefore, ds = du/T = C dT/T

2 2

2

2 1

1

1 1

avg

Integrate to determine the entropy change during a process

s

( )

ln( )

where C is the averaged specific heat of the substance

over the given temperature range

avg

dT

T

s

ds

C T

C

T

T

 









• Specific heats for some common liquids and solids can be found in thermodynamic tables such as Table A-14 to A-18

Example – Entropy Change of Incomp. Substances

A 1-kg metal bar, initially at 1000 K, is removed from an oven and quenched by immersing in a closed tank containing 20 kg of water, initially at 300 K. Assume both substances are incompressible and C_water= 4(kJ/kg K), C_metal = 0.4(kJ/kg K). Neglect heat transfer between the tank and its surroundings. (a) Determine the final temperature of the metal bar, (b) entropy generation

during the process.

T

=1000 K, m

=1kg,

c

=0.4 kJ/kg K

Solution

water metal bar f

(a) Energy balance from the first law:

U Q-W 0, no heat transfer and no work done

U

U

0, both bar and water reach final temperature T

(

)

(

) 0

( /

)

( /

w w f w m m m f

w w m w m m f

w w

m c T

T

m c T

T

m c c T

m T

T

m c c

 





 















(20)(10)(300) (1)(1000)

303.5( )

)

(20)(10) 1

m m

K

m











)

/

(

451

.

0

477

.

0

928

.

0

1000

5

.

303

ln

)

4

.

0

)(

1

(

300

5

.

303

ln

)

4

)(

20

(

ln

ln

s(bar)

s(water)

s

on)

s(generati

s

system

the

of

balance

entropy

the

0,

Q

outside

he

fer with t

heat trans

No

(b)

K

kJ

s

T

T

c

m

T

T

c

m

s































Entropy Change of Ideal Gases

(YAC:6-9)

From Gibbs’ equations, the change of entropy of an ideal gas can

be expressed as:

dP

T

v

T

dh

dv

T

P

T

du

ds









For an ideal gas, u=u(T) and h=h(T), du=c

(T)dT and dh=c

(T)dT and

Pv = RT

2 2

2 2

2 1 2 1

1 1

1 1

p v

( )

,

( )

By integration, the change of the entropy is

( )

ln( ) or

( )

ln( )

we need to know the function c (T) and c (T) in order to compl

v P

v P

dT

dv

dT

dP

ds c T

R

and ds c T

R

T

v

T

P

dT

v

dT

P

s

s

c T

R

s

s

c T

R

T

v

T

P









 





 





Entropy Change of Ideal Gases – Special Cases

However, for certain cases, one can make simplifying assumptions.

Case 1: If specific heats are assumed constant, integration is simplified:

Note: The above is strictly true for monatomic gases, e.g.He, Ar

It is fairly accurate if the temperature difference is small.

Case 2: Calculate the specific heat at an average temperature, T_avg, and assume it to be constant. This also allows the specific heat to be taken out of the integral

Note:

This approximation is generally fairly accurate if the temperature difference is not too large, usually good if T < few hundred degrees.

Strictly speaking, one should look at the temp. dependence of specific heats for the particular substance to evaluate the validity of this approximation.

2 2

2 1

1 1

2 2

2 1

1 1

ln( )

ln( ) or

ln( )

ln( )

v

P

T

v

s

s

c

R

T

v

T

P

s

s

c

R

T

P

 



Isentropic Processes for Ideal Gases

If a process is isentropic (that is adiabatic and reversible), ds = 0, s₁=s₂, then it can be shown that:

 1 /

1

2 1 2 2

1 2 1 1

p

2 1

1 2 v

k

( ) , and

( )

c

and

( ) , where

c

k k

k

k

T

v

T

P

T

v

T

P

P

v

P

v













The above are referred to as the: first, second and third, respectively, isentropic relations for Ideal Gases (assuming constant specific heats).

They can also be written as:

Tvk-1 _{= constant First isentropic relation}

TP(1-k)/k _{= constant Second isentropic relation}

Example

• Air is compressed from an initial state of 100 kPa and 300

K to 500 kPa and 360 K. Determine the entropy change

using constant c

=1.003 (kJ/kg K)

2 2

2 1 P

1 1

2 1

ln( )

ln( ) if c is constant

360

500

1.003ln

(0.287) ln

0.279(

/

)

300

100

P

T

P

s

s

c

R

T

P

s

s

kJ kg K

 



 



 